(પરમાણ્વીય ક્રમાંક : $Ti = 22,\,Cr = 24,$ $ Mn = 25,\,Ni = 28$)
Here, $n=$ number of unpaired electrons
$\Rightarrow 2.83=\sqrt{n(n+2)}$
$\Rightarrow(2.83)^{2}=n(n+2)$
$0=8.00+n^{2}+2 n$
$n^{2}+ 2n-8=0$
$(n+4)(n-2)=0$
$n=2$ $n=-\,4$
$-\,4$ not be consider
Hence, $Ni^{2+}$ possesses a magnetic moment of $2.83\,\mathrm{B} . \mathrm{M}$